Question: $ g(x) = \int_{2}^{x}(10t + 2)\,dt\,$ $ g\,^\prime(3)\, =$
The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = 10t +2$ is continuous on $[2,3]$. Applying the theorem We're given: $ g(x) = \int_{2}^{x}(10t + 2)\,dt\,$ So the theorem tells us: $ g\,^\prime(x) =10x + 2$ Evaluating $g'(3)$ $ g'(3)= 10(3) + 2 = 32$ The answer: $g'(3)=32$